Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main(){
  int n; cin >> n;
  vector<long double> f(n + 1);
  f[0] = 1;
  for(int i = 0; i < n; i++){
    long double p0; cin >> p0;
    for(int j = n - 1; j >= 0; j--){
      f[j + 1] += p0 * f[j];
      f[j] *= (1 - p0);
    }
  }
  long double pr = 0;
  for(int i = (n + 1)/2; i <= n; i++) pr += f[i];
  cout << fixed << setprecision(12) << pr << endl;
  return 0;
}